2005 AMC 10B Problem 17

Each equation turns one number into a power of the number before it. So we can chain them together, starting from $8$ and working back to $4$.

1. Start with $8 = 7^d$.
2. The previous equation says $7 = 6^c$. Replace $7$:
$$8 = (6^c)^d = 6^{cd}.$$
(A power of a power multiplies the exponents.)
3. Replace $6$ with $5^b$:
$$8 = (5^b)^{cd} = 5^{bcd}.$$
4. Replace $5$ with $4^a$:
$$8 = (4^a)^{bcd} = 4^{abcd}.$$
5. Now write both sides as powers of $2$. Since $8 = 2^3$ and $4 = 2^2$:
$$2^3 = (2^2)^{abcd} = 2^{2abcd}.$$
6. Same base, so the exponents must match: $2abcd = 3$, which gives $abcd = \frac{3}{2}$.

The answer is $\boxed{\textbf{(B)}\ \frac{3}{2}}$.

When the problem asks for a product like $abcd$ rather than each letter, look for a way to make the product appear on its own. Here, plugging each equation into the next stacks the exponents into one product. The middle numbers $5$, $6$, $7$ drop out, and only the first base $4$ and the last number $8$ matter.

With logarithms: $a = \log_4 5$, $b = \log_5 6$, $c = \log_6 7$, $d = \log_7 8$. Using the change-of-base rule, each one is a fraction of logs, and multiplying them cancels the $\log 5$, $\log 6$, $\log 7$ top and bottom. What remains is $\frac{\log 8}{\log 4} = \frac{3 \log 2}{2 \log 2} = \frac{3}{2}$.